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3 years ago

How is capacitance related to the distance between the plates of a capacitor?

Physics

1 answer:

svetlana [45]3 years ago

5 0

Answer:C

Explanation: I studied, and C is correct

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What do you understand if the frequency of a sound is 70 Hz?

ankoles [38]

Answer:

see below

Explanation:

frequency of a wave is the number of complete waves produced per second

so 70 Hz

means, that 70 complete waves are produced in one second,

hope this helps

4 0

3 years ago

If the distance between the bounces were 1.95 m instead of 1.30 m, but the height remained at 1.30 m, which of your answers to P

Oksanka [162]

Answer:

Below is the calculation of all the required quantities. From the calculations, Vx must change, the velocity before hitting the ground must change and the angles also must change.

Explanation:

Below is an attachment that contains the calculation steps.

Thank you for reading.

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3 years ago

POINTS + BRAINLIEST TO CORRECT ANSWER

Tpy6a [65]

Answer:kahhot.it

Explanation:hghh

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4 years ago

Which statement is best supported by the information in the chart? Wave X and Wave Y are mechanical waves, and Wave Z is an elec

cluponka [151]

Answer:

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3 years ago

A 1-mCi source of^60 Co is placed in the center of a cylindrical water-filled tank with an inside diameter of 20 cm and depth of

LenKa [72]

To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.

The energy fluency is given by the equation

$\Psi=4RcE\pi$

Where

$\Psi =$ The energy fluency

c = Activity of the source

r = distance

E = electric field

In the other hand we have the equation for current in materials, which is given by

$I= I_0 e^{-\mu_{h20}X_{h2o}} e^{-\mu_{Fe}X_{Fe}}$

Then replacing our values we have that

$I = 1*10^{-3} * 3.3*10^{10} * e ^{-0.06*1.1} e^{-0.058*7.861}$

$I = 1.3*10^7 Bq$

We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.

Now the energy fluency would be,

$\Psi = \frac{cE}{4\pir^2}$

$\Psi = \frac{1.3*10^7*2*1.25}{4\pi*11^2}$

$\Psi = 2.14*10^4 MeV/cm^2.s$

The uncollided flux density at the outer surface of the tank nearest the source is $\Psi = 2.14*10^4 MeV/cm^2.s$

6 0

3 years ago